The document provides data from a test on a flat-walled furnace with inner and outer brick walls of unknown conductivity. The temperatures at various points within the furnace walls are given. Using this data and the thickness of each material, the conductivity of each brick type is calculated. The percentage of heat loss saved by adding 5cm of magnesia insulation to the furnace is calculated to be 52.73%.

1. CONDUCTION Problem Set Alejandro, Jan Rannel Granada, Gat Hendrix, Jamie Salcedo, Angel Frances Villaflor, Kaye

2. STEADY STATE APPLICATIONS 5. The following data was obtained in a test on a flat-walled furnace, the linings of which consist of a 11.5 cm. non-corrosive brick of unknown conductivity and the outer wall of 20 cm clay brick, also of unknown conductivity. The temperature of the inner wall (flame side) was found to be 735°C and that of the outer wall 185°C. This furnace was lagged with 5 cm of magnesia (k=0.07 W/mK) thermocouples inserted at various points and the ff. data taken: Calculate the percentage of heat loss that is saved by the lagging. 88°C Temp of the outer surface of magnesia 475°C Temp at the junction of ordinary brick and magnesia 700°C Temp at the junction of brick layers 735°C Temp of inner wall (flame side)

3. STEADY STATE APPLICATIONS Magnesia

4. Solution: Basis: Am = 1m2 From the data of composition with magnesia: q = ∆T/R R = ∆x/(km)(Am) q = ∆T(km)(Am)/ ∆x Series: q1 = q2 = q3 =qtotal q of magnesia(q3) = (475-88)K(0.07W/mK)( 1m2) = 541.8 W 0.05m q of magnesia = q of non-corrosive brick = q of clay brick = q total = 541.2 W km = q(∆x)/(∆T)(Am) km of non-corrosive brick = (541.8W)(0.115m)/(735-700)( 1m2) = 1.7802 W/mK km of clay brick = (541.8 W)(0.20m)/(700-475) ( 1m2) = 0.4816 W/mK non-corrosive brick clay brick Magnesia 735°C 700°C 475°C 88°C

5. Using the data in the composition without the magnesia: Series: Rtotal = R1 + R2 + R3… Getting the R: R = ∆x/(km)(Am) R non-corrosive brick = (0.115m) /(1.7802 W/mK) ( 1m2) = 0.06460 K/W R of clay brick = (0.20m)/(0.4816 W/mK)( 1m2) = 0.4153 K/W Rtotal = 0.4799K/W q total = (735-185)K/(0.4799 K/W) = 1146.07 W percentage of heat loss that is saved by the lagging: (q total without the magnesia – q total with magnesia) x 100% q total without magnesia (1146.07W-541,8W) x 100% = 52.73% 1146.07W non-corrosive brick clay brick 735°C 185°C

8. UNSTEADY STATE APPLICATIONS Solution: h = 204 _kJ_ x 1000J x 1h___ = 56.67 W/m2K hm2K 1 kJ 3600 s X1 = 2.5/100 = 0.0125m 2 Y = 315 – 93 = 0.74 315 – 16 m = k__ = 38.0 W/mK__________ = 53.64 hX1 56.67 W/m2K (0.0125m) From the Heisler Chart(Cylinder): X = 10.1 = αt__ = (0.186) t t = 0.0085s X12 (0.0125)2